Section 2: Operations on Functions · Section 2: Operations on Functions Composition of Functions Suppose we wanted to calculate how much it costs to heat a house on a particular

Chapter 1 Review Applied Calculus 19

This chapter was remixed from Precalculus: An Investigation of Functions, (c) 2013 David Lippman and Melonie

Rasmussen. It is licensed under the Creative Commons Attribution license.

Section 2: Operations on Functions

Composition of Functions Suppose we wanted to calculate how much it costs to heat a house on a particular day of the year.

The cost to heat a house will depend on the average daily temperature, and the average daily

temperature depends on the particular day of the year. Notice how we have just defined two

relationships: The temperature depends on the day, and the cost depends on the temperature.

Using descriptive variables, we can notate these two functions.

The first function, C(T), gives the cost C of heating a house when the average daily temperature

is T degrees Celsius, and the second, T(d), gives the average daily temperature of a particular city

on day d of the year. If we wanted to determine the cost of heating the house on the 5th

day of

the year, we could do this by linking our two functions together, an idea called composition of

functions. Using the function T(d), we could evaluate T(5) to determine the average daily

temperature on the 5th

day of the year. We could then use that temperature as the input to the

C(T) function to find the cost to heat the house on the 5th

day of the year: C(T(5)).

Composition of Functions

When the output of one function is used as the input of another, we call the entire

operation a composition of functions. We write f(g(x)), and read this as “f of g of x” or “f

composed with g at x”.

An alternate notation for composition uses the composition operator:

))(( xgf is read “f of g of x” or “f composed with g at x”, just like f(g(x)).

Example 1

Suppose c(s) gives the number of calories burned doing s sit-ups, and s(t) gives the number of

sit-ups a person can do in t minutes. Interpret c(s(3)).

When we are asked to interpret, we are being asked to explain the meaning of the expression

in words. The inside expression in the composition is s(3). Since the input to the s function is

time, the 3 is representing 3 minutes, and s(3) is the number of sit-ups that can be done in 3

minutes. Taking this output and using it as the input to the c(s) function will gives us the

calories that can be burned by the number of sit-ups that can be done in 3 minutes.

Composition of Functions using Tables and Graphs

When working with functions given as tables and graphs, we can look up values for the functions

using a provided table or graph. We start evaluation from the provided input, and first evaluate

the inside function. We can then use the output of the inside function as the input to the outside

function. To remember this, always work from the inside out.


Example 2

Using the graphs below, evaluate ( (1))f g .

g(x) f(x)

To evaluate ( (1))f g , we again start with the inside evaluation. We evaluate (1)g using the

graph of the g(x) function, finding the input of 1 on the horizontal axis and finding the output

value of the graph at that input. Here, (1) 3g . Using this value as the input to the f

function, ( (1)) (3)f g f . We can then evaluate this by looking to the graph of the f(x)

function, finding the input of 3 on the horizontal axis, and reading the output value of the

graph at this input. Here, (3) 6f , so ( (1)) 6f g .

Composition using Formulas

When evaluating a composition of functions where we have either created or been given

formulas, the concept of working from the inside out remains the same. First we evaluate the

inside function using the input value provided, then use the resulting output as the input to the

outside function.

Example 3

Given tttf 2)( and 23)( xxh , evaluate ( (1))f h .

Since the inside evaluation is (1)h we start by evaluating the h(x) function at 1:

52)1(3)1( h

Then ( (1)) (5)f h f , so we evaluate the f(t) function at an input of 5:

2055)5())1(( 2 fhf

We are not limited, however, to using a numerical value as the input to the function. We can put

anything into the function: a value, a different variable, or even an algebraic expression,

provided we use the input expression everywhere we see the input variable.


Example 4

Let 2)( xxf and xx

xg 21

)( , find f(g(x)) and g(f(x)).

To find f(g(x)), we start by evaluating the inside, writing out the formula for g(x)

xx

xg 21

)(

We then use the expression 1

2xx

as input for the function f.

x

xfxgf 2

1))((

We then evaluate the function f(x) using the formula for g(x) as the input.

Since 2)( xxf then

2

21

21

x

xx

xf

This gives us the formula for the composition:

2

21

))((

x

xxgf

Likewise, to find g(f(x)), we evaluate the inside, writing out the formula for f(x)

2))(( xgxfg

Now we evaluate the function g(x) using x2 as the input.

2

22

1))(( x

xxfg

Example 5

A city manager determines that the tax revenue, R, in millions of dollars collected on a

population of p thousand people is given by the formula pppR 03.0)( , and that the

city’s population, in thousands, is predicted to follow the formula 23.0260)( tttp ,

where t is measured in years after 2010. Find a formula for the tax revenue as a function of

the year.

Since we want tax revenue as a function of the year, we want year to be our initial input, and

revenue to be our final output. To find revenue, we will first have to predict the city

population, and then use that result as the input to the tax function. So we need to find

R(p(t)). Evaluating this,

222 3.02603.026003.03.0260))(( ttttttRtpR

This composition gives us a single formula which can be used to predict the tax revenue

during a given year, without needing to find the intermediary population value.


For example, to predict the tax revenue in 2017, when t = 7 (because t is measured in years

after 2010)

079.12)7(3.0)7(260)7(3.0)7(26003.0))7(( 22 pR million dollars

Later in this course, it will be desirable to decompose a function – to write it as a composition of

two simpler functions.

Example 6

Write 253)( xxf as the composition of two functions.

We are looking for two functions, g and h, so ))(()( xhgxf . To do this, we look for a

function inside a function in the formula for f(x). As one possibility, we might notice that 25 x is the inside of the square root. We could then decompose the function as:

25)( xxh

xxg 3)(

We can check our answer by recomposing the functions:

22 535))(( xxgxhg

Note that this is not the only solution to the problem. Another non-trivial decomposition

would be 2)( xxh and xxg 53)(

Transformations of Functions Transformations allow us to construct new equations from our basic toolkit functions. The most

basic transformations are shifting the graph vertically or horizontally.

Vertical Shift

Given a function f(x), if we define a new function g(x) as

( ) ( )g x f x k , where k is a constant

then g(x) is a vertical shift of the function f(x), where all the output values have been

increased by k.

If k is positive, then the graph will shift up

If k is negative, then the graph will shift down

Horizontal Shift


( ) ( )g x f x k , where k is a constant

then g(x) is a horizontal shift of the function f(x)

If k is positive, then the graph will shift left


If k is negative, then the graph will shift right

Example 7

Given ( )f x x , sketch a graph of 313)1()( xxfxh .

The function f is our toolkit absolute value function. We know that this graph has a V shape,

with the point at the origin. The graph of h has transformed f in two ways: ( 1)f x is a

change on the inside of the function, giving a horizontal shift left by 1, then the subtraction by

3 in ( 1) 3f x is a change to the outside of the function, giving a vertical shift down by 3.

Transforming the graph gives

Example 8

Write a formula for the graph shown, a transformation of

the toolkit square root function.

The graph of the toolkit function starts at the origin, so this

graph has been shifted 1 to the right, and up 2. In function

notation, we could write that as ( ) ( 1) 2h x f x . Using

the formula for the square root function we can write

( ) 1 2h x x

Note that this transformation has changed the domain and

range of the function. This new graph has domain [1, )

and range [2, ) .

Another transformation that can be applied to a function is a reflection over the horizontal or

vertical axis.


Reflections

Given a function f(x), if we define a new function g(x) as ( ) ( )g x f x ,

then g(x) is a vertical reflection of the function f(x), sometimes called a reflection about

the x-axis

If we define a new function g(x) as ( ) ( )g x f x ,

then g(x) is a horizontal reflection of the function f(x), sometimes called a reflection about

the y-axis

Example 9

A common model for learning has an equation similar to

( ) 2 1tk t , where k is the percentage of mastery that can be

achieved after t practice sessions. This is a transformation of

the function ( ) 2tf t shown here. Sketch a graph of k(t).

This equation combines three transformations into one equation.

A horizontal reflection: ( ) 2 tf t combined with

A vertical reflection: ( ) 2 tf t combined with

A vertical shift up 1: ( ) 1 2 1tf t

We can sketch a graph by applying these transformations one at a time to the original

function:

The original graph Horizontally reflected Then vertically reflected

Then, after shifting up 1, we get the final graph:

( ) ( ) 1 2 1tk t f t .

Note: As a model for learning, this function

would be limited to a domain of 0t , with

corresponding range [0,1) .


With shifts, we saw the effect of adding or subtracting to the inputs or outputs of a function. We

now explore the effects of multiplying the outputs.

Vertical Stretch/Compression


)()( xkfxg , where k is a constant

then g(x) is a vertical stretch or compression of the function f(x).

If k > 1, then the graph will be stretched

If 0< k < 1, then the graph will be compressed

If k < 0, then there will be combination of a vertical stretch or compression with a vertical

reflection

Example 10

The graph to the right is a transformation of the toolkit

function 3)( xxf . Relate this new function g(x) to f(x),

then find a formula for g(x).

When trying to determine a vertical stretch or shift, it is

helpful to look for a point on the graph that is relatively

clear. In this graph, it appears that 2)2( g . With the

basic cubic function at the same input, 82)2( 3 f .

Based on that, it appears that the outputs of g are ¼ the

outputs of the function f, since )2(4

1)2( fg . From this

we can fairly safely conclude that:

)(4

1)( xfxg

We can write a formula for g by using the definition of the function f

3

4

1)(

4

1)( xxfxg

Combining Transformations

When combining vertical transformations, it is very important to consider the order of the

transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does

not create the same graph as vertically stretching by 2 and then vertically shifting by 3. The order

follows nicely from order of operations.

Combining Vertical Transformations

When combining vertical transformations written in the form kxaf )( ,


first vertically stretch by a, then vertically shift by k.

Example 11

Write an equation for the transformed graph of the

quadratic function shown.

Since this is a quadratic function, first consider what the

basic quadratic tool kit function looks like and how this

has changed. Observing the graph, we notice several

transformations:

The original tool kit function has been flipped over the x

axis, some kind of stretch or compression has occurred,

and we can see a shift to the right 3 units and a shift up 1

unit.

In total there are four operations:

Vertical reflection, requiring a negative sign outside the function

Vertical Stretch

Horizontal Shift Right 3 units, which tells us to put x-3 on the inside of the function

Vertical Shift up 1 unit, telling us to add 1 on the outside of the function

By observation, the basic tool kit function has a vertex at (0, 0) and symmetrical points at

(1, 1) and (-1, 1). These points are one unit up and one unit over from the vertex. The new

points on the transformed graph are one unit away horizontally but 2 units away vertically.

They have been stretched vertically by two.

Not everyone can see this by simply looking at the graph. If you can, great, but if not, we can

solve for it. First, we will write the equation for this graph, with an unknown vertical stretch. 2)( xxf The original function

2)( xxf Vertically reflected 2)( axxaf Vertically stretched

2)3()3( xaxaf Shifted right 3

1)3(1)3( 2 xaxaf Shifted up 1

We now know our graph is going to have an equation of the form 1)3()( 2 xaxg . To

find the vertical stretch, we can identify any point on the graph (other than the highest point),

such as the point (2,-1), which tells us 1)2( g . Using our general formula, and

substituting 2 for x, and -1 for g(x)

a

a

a

a

2

2

11

1)32(1 2

This tells us that to produce the graph we need a vertical stretch by two.


The function that produces this graph is therefore 1)3(2)( 2 xxg .

Example 12

On what interval(s) is the function

31

2)(

2

xxg increasing and decreasing?

This is a transformation of the toolkit reciprocal squared function, 2

1)(

xxf :

2

2)(2

xxf

A vertical flip and vertical stretch by 2

21

2)1(2

xxf A shift right by 1

3

1

23)1(2

2

xxf A shift up by 3

The basic reciprocal squared function is increasing on )0,( and

decreasing on ),0( . Because of the vertical flip, the g(x)

function will be decreasing on the left and increasing on the right.

The horizontal shift right by 1 will also shift these intervals to the

right one. From this, we can determine g(x) will be increasing on

),1( and decreasing on )1,( . We also could graph the

transformation to help us determine these intervals.

1.2 Exercises

Given each pair of functions, calculate 0f g and 0g f .

1. 4 8f x x , 27g x x 2. 5 7f x x , 24 2g x x

3. 4f x x , 312g x x 4. 1

2f x

x

, 4 3g x x

Use the table of values to evaluate each expression

5. ( (8))f g

6. 5f g

7. ( (5))g f

8. 3g f

9. ( (4))f f

10. 1f f

11. ( (2))g g

12. 6g g

x ( )f x ( )g x

0 7 9

1 6 5

2 5 6

3 8 2

4 4 1

5 0 8

6 2 7

7 1 3

8 9 4

9 3 0


Use the graphs to evaluate the expressions below.

13. ( (3))f g

14. 1f g

15. ( (1))g f

16. 0g f

17. ( (5))f f

18. 4f f

19. ( (2))g g

20. 0g g

For each pair of functions, find f g x and g f x . Simplify your answers.

21. 1

6f x

x

,

76 g x

x 22.

1

4f x

x

,

24g x

x

23. 2 1f x x , 2g x x 24. 2f x x , 2 3g x x

25. f x x , 5 1g x x 26. 3f x x , 3

1xg x

x

27. If 4 6f x x , ( ) 6 g x x and ( ) h x x , find ( ( ( )))f g h x

28. If 2 1f x x , 1

g xx

and 3h x x , find ( ( ( )))f g h x

29. The function ( )D p gives the number of items that will be demanded when the price is p. The

production cost, ( )C x is the cost of producing x items. To determine the cost of production

when the price is $6, you would do which of the following:

a. Evaluate ( (6))D C b. Evaluate ( (6))C D

c. Solve ( ( )) 6D C x d. Solve ( ( )) 6C D p

30. The function ( )A d gives the pain level on a scale of 0-10 experienced by a patient with d

milligrams of a pain reduction drug in their system. The milligrams of drug in the patient’s

system after t minutes is modeled by ( )m t . To determine when the patient will be at a pain

level of 4, you would need to:

a. Evaluate 4A m b. Evaluate 4m A

c. Solve 4A m t d. Solve 4m A d


Find functions ( )f x and ( )g x so the given function can be expressed as h x f g x .

31. 2

2h x x 32. 3

5h x x

33. 3

5h x

x

34.

2

4

2h x

x

35. 3 2h x x 36. 34h x x

Sketch a graph of each function as a transformation of a toolkit function.

37. 2( 1) 3f t t 38. 1 4h x x

39. 3

2 1k x x 40. 3 2m t t

41. 2

4 1 5f x x 42. 2

( ) 5 3 2g x x

43. 2 4 3h x x 44. 3 1k x x

Write an equation for each function graphed below.

45. 46.

47. 48.

49. 50.


For each function graphed, estimate the intervals on which the function is increasing and

decreasing.

51. 52.

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Section 2: Operations on Functions · Section 2: Operations on Functions Composition of Functions Suppose we wanted to calculate how much it costs to heat a house on a particular